Prove HK=KH iff HK is a Subgroup of G

  • Thread starter Thread starter annoymage
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving that the product of two subgroups, HK and KH, is equal if and only if HK is a subgroup of G. The participants demonstrate that if HK is a subgroup, then KH is a subset of HK, and they successfully derive that for any element x in HK, its inverse x^-1 can be expressed as k^{-1}h^{-1}, which belongs to KH. This establishes the necessary condition for HK being a subgroup of G.

PREREQUISITES
  • Understanding of subgroup properties in group theory
  • Familiarity with the notation and operations of group elements
  • Knowledge of inverse elements in group theory
  • Basic experience with mathematical proofs
NEXT STEPS
  • Study the properties of subgroup products in group theory
  • Learn about the criteria for subgroup tests in abstract algebra
  • Explore the concept of normal subgroups and their significance
  • Investigate the implications of subgroup intersections and unions
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and anyone interested in understanding subgroup relationships and their proofs.

annoymage
Messages
360
Reaction score
0

Homework Statement



if H and K are arbitrary subgroup of G, prove that HK=KH iff HK is a subgroup of G

Homework Equations



n/a

The Attempt at a Solution



no problem to prove => direction

for <= i can prove KH is a subset of HK

only i got troubled to show HK ia subset of KH

x in HK

x=hk for some h in H ,k in K

i manipulate it many ways and always got the form x=khk for some h in H ,k in K

HELP, and sorry no latex ,i'm very buzy now ;P
 
Physics news on Phys.org


Take b in HK so that b-1 = hk is in HK. How do you get b back?
 


so k^{-1}h^{-1} \in KH[/tex] <br /> <br /> and hk \in HK, what&#039;s its inverse?
 


aahhhh i see,

b in HK

b=hk for some h in H k in K

b^{-1} also is in HK

imply
<br /> b^{-1}=(hk)^{-1}=k^{-1}h^{-1} \in KH <br />

right ??
 


wait wrong,

for any x in HK, x^-1 in HK so x^-1=hk for some h and k

then <br /> <br /> x=(x^{-1})^{-1}=(hk)^{-1}=k^{-1}h^{-1} \in KH<br /> <br />

now this is correct right?
 


yeah that looks good
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
13K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K